CISCO PLACEMENT PAPER 3

CISCO IIT ROORKEE PLACEMENT PAPER AND HR INTERVIEW QUESTIONS

NOTE : Here are two papers H/W and S/W both are compulsary for all .
they allowed all E&c , computers and electrical (both ME and BE).
and there is no Aptitude test .
H/W 30 bits
S/W 30 bits total time is 60 minuts.

1> The starting location of an array is 1000. If the array[1..5/...4] is
stored in row major order, what is the location of element [4,3]. Each work
occupies 4 bytes. Ans: 1068 or 1056

2> In a 3 ary tree, which has 3childs for every node,If the number of
internal nodes are N, then the total number of leaf nodes.
ANS: 2^(N-1) +3

3> The locality of reference means...

4> If two ausigned 8 bit numbers are multiplied what is the memory space required........... Ans: 64k x 16 bits

5> The vector address of RST 7.5 is ......

ANS: 003C (multiply 7.5 by 8 and convert to hex)

6> char a =0xAA;
`int b;
b= (int) a;
b=b>>4;
printf("%x",b);
What is the output of the above program
a)0x0000000a
b)0x000000aa
c)0x00000aa0
d)0xfffffffa

7> struct s1 { struct { struct {int x;}s2}s3}y;
How to access x?
Ans : Illelgeal strucrure operation.

8> Why there is no recursion in Fortran? *******
ANS: There is no dynamic allocation.

9> What is the worst case complexity of Quick sort?
ANS: O(n^2)

10)What will be sequence of o/s activities when an interrupt occurs.
Ans:resumes the current process after servicing the interupt.

11> In a sequential search, the average no. of comparisons it takes to search through n elements is

Ans: (n+1)/2.

12> What is the size of the array declared as double * X[5]

ANS: 5*sizeof (double *)

13> A binary search tree is given and asked to write the preorder traversal result. 12345678
Ans : 53124768

14> If size of the physical memory is 2^32-1, then the size of virtual
memory...... (doubt ..2^32.)

15> S-> A0B
A-> BB0
B-> AA1 How many strings of length 5 are possible with the above
productions?

16> (3*4096+15*256+3*16+3). How many 1's are there in the binary
representation
of the result.
ANS: 10

17> In memory mapped I/O how I/O is accessed....

ANS: Just like a memory location (Means, I/O devices can be accessed
using the instructions like mov A,M etc...)

18> What is the use of ALE in 8085..
ANS: To latch the lower byte of the address.

19> If the logical memory of 8 X 1024 is mapped into 32 frames, then the number of bits for the logical address ......
ANS: 13

20> Context free grammar is useful for...
ANS: If-then structures.

21> In ternary number representation, numbers are represented as 0,1,-1.
Here -1 is represented as - (1 bar). Then how is 352/9 represented......
1

22> There are processors which take 4,1,8,1 machine cycles respectively. If these are executed in round robin fashion with a time quantum of 1, what is the time it take for process 4 to complete....
ANS: 9

23> The minimum frequency of operation is specified for every processor because......
a)for interfacing slow peripherals
b)dynamic memory refreshing.
c)to make compatible with other processor.

24) linked list implementation , the following search is not applicable:
Ans: binary search.

25> The reason for preferring CMOS over NMOS is....
Ans: Low power consumption.

26> Two's complement of binary numbers A,B are given and asked to find out A-B.
Ans: 00100

27> Each character is represented by 7 bits, 1 bit is used to represent error bit and another bit for parity. If total number of bits transmitted is 1200bits, then number of symbols that can be transmitted.......
Ans: 133

28> One question about the setassociativity of cache..
Ans:d 4,6

29> Write the postfix form of the following expression...
A+[[(B+C)+(D+E)*F]/G] Ans : abc+de+f*+g/+

30> What is the function of the linker......

31> void f(int y)
{
struct s *ptr;
ptr = malloc (sizeof (struct)+99*sizeof(int));
}

struct s{
int i;
float p;
};

when free(ptr) is executed, then what will happen??

32) To concatenate twe linked lists strings, the order is 0(1) is for
Ans:circular doublly linked list ( please check it).


HARDWARE PAPER.

bOTH S/W AND S/W are compulsory.


1. to find out stack fault of a 3 input nand gate how many necessary
input vectors are needed
2. solving k_map Ans: b+c

3.parity generation i.e even parity
ANs : c

4. nand gate becomes --- gate for negative logic Ans: OR

5.Advantage of cmos over nmos : ans: Low power dessipation

6.Adv of syncronous ckts on async. ckts

7. 5 problems on addressing of memories

8.Function of ale in 8085

10.A voice signal sample is stored as one byte. Frequency range is 20Hz to 16hz. What is the memorysize required to store 4 minutes voice signal?
Ans 15MB

11)What will the the controller do before interrupting CPU?
ANs : stores the data into its memory and then interrupts the CPU. nd other are digital fundamentls

12) In a normalised floating point representation, mantessa is represented by 24 bits and exponent with 8 bits using signed representation. What is range?

14) stack uses Ans: LIFO

15) Where will be the actual address of the subroutine is placed for vectored interrupts?
Ans: Fixed location.

16) Equivalent Gray code reprasentation of AC2H.

CONCENTRATE ON DIGITAL LOGIC( GATES, RAM ADDRESSING, NO OF BITS REQUIRED,)